Let A=${a_n A=$\{a_n : n\in \omega }\subset \}\subset 2^{\omega\times\omega}$ be nonempty countable without isolate point(i.e Homeomorphism isolated points (i.e. homeomorphic to Q), $\mathbb{Q}$), and satisfy $ \forall n\in \omega \exists^\infty m|{k:a_n(m,k)=1}|=\omega m|\{k:a_n(m,k)=1\}|=\omega $. Is Does there exist $ a\in cl(A)\setminus A$ satisfy satisfying $\exists^\infty m|{k:a(m,k)=1}|=\omega m|\{k:a(m,k)=1\}|=\omega $?
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Let $A={a_n:n\in A=${a_n : n\in \omega }\subset 2^{\omega\times\omega}$ be nonempty countable without isolate point(i.e Homeomorphism to Q), and satisfy $ \forall n\in \omega \exists^\infty m|{k:a_n(m,k)=1}|=\omega $. Is there exist $ a\in cl(A)\setminus A$ satisfy $\exists^\infty m|{k:a(m,k)=1}|=\omega $? |
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Let $A=(a_n:n\in A={a_n:n\in \omega )}\subset 2^{\omega\times\omega}$ be nonempty countable without isolate point(i.e Homeomorphism to Q), and satisfy $ \forall n\in \omega \exists^\infty m|(k:a_n(m,k)=1)|=\omega m|{k:a_n(m,k)=1}|=\omega $. Is there exist $ a\in cl(A)\setminus A$ satisfy $\exists^\infty m|(k:a(m,k)=1)|=\omega m|{k:a(m,k)=1}|=\omega $? |
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